# the position of a body moving along x-axis at time t is given by x = (t2-4t+6) m . The distance travelled by body in time interval t = 0 to t = 3 s is -(1) 5 m(2) 7m(3) 4 m(4) 3 m

what to find in this??

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distance travelled in time interval t = 0 to t = 3s

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3m

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3m is displacement it is also the distance if body does not change its direction.

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it's answer is 5m according to the aakash booklet but i don't know how.

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Exactly the answer is 5m.... But how
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Here x=t^2-4t+6 equation represent the position of that particle along x axis at given time t. Now at t=0 , particle present at x=6 point means x(6,0) then t=1 ,x(3,1); then t=2,x(2,2); and last t=3,x(3,3). Now distance is sclar quantity so along x axis particle move the distance x(6,0) to x(3,1) = 3m, x(3,1) to x(2,2) =1m and last the particle move x(2,2) to x(3,1)=1m . So the distance travelled by body in time interval t=0 to t=3s is (3+1+1)=5m
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As already told by Abu syed. x at 0 sec = 0^2-4*0+6=6m x at 1 sec = 3m x at 2 sec = 2m x at 3 sec = 3m Now let's take out distance at particular second. x at 1st sec = 3m x at 2nd sec (x at 2 sec - x at 1sec) = 2-3 = 1m (as distance is scalar quantity and hence has no direction therefore no-ve sign) x at 3rd sec (x at 3 sec - x at 2sec) = 3-2 = 1m. Now add all the individual distances I.e. 3+1+1 = 5m.
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North kolkata

• -14
5 meter
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x=t2​ - 4t +6
dx/dt= 2t - 4
at t=2, v= 0 so the body revers its position so,
xt=0 = 6m     xt=2s​ = 2m     xt=3s = 3m
the distance = (6- 2)+ (3- 2) = 5m
displacement = 3m

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Answer is correct but I think so that this procedure is correct
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x=t^2-4t+6
at t=0 x=0 so,
(3)^2 -4*3+6=9-12+6
=3
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Yes answer is 5 but I am not able to understand
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Please draw diagram of this ques
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Graphical equations of motion in class 9th please tell me pleaae
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And..
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X= t- 4t+6

dx/dt= 2t-4 =0 (where v=0)

t=2 v will be 0 ie body will turn back.

so for 1st 2 secs x= 4-8+6= 2m

for last 1  sec x= 1-4+6=3m

adding 2+3= 5 so the body travels 5m.

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Hope this helps, sry for bad handwriting. • 54
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Ans is 7 in aakash booklet
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Saththiyama purila pa
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(1) 5 m
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Answer is 3m (d) part because from t = 0 to t= 3 sec
x = t^2 - 4t + 6
putting time t = 0 in this equation
the anser will equal to   = 0 * 0 - 4 * 0 + 6 = 6 m
putting t= 3 in this equation
3*3  - 4 * 3 + 6 = 9 - 12 + 6 = 15 - 12 = 3m

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Answer is 5m as the body is taking turn at t=2s • 8
Differentiate wrt time to get velocity and then check whether it is zero at any time or not and. Then apply the formula
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4.3m
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5m/s
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Ans-3m
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Answer is likely to be 2
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5 m is the answer bro

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x = t2 -4t +6
Putting t= 3
So x = 32- 4*3+6
9-12+6=3m
So displacement = 3m
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The answer of this question has been given below in image • -13
Abu syed. x at 0 sec = 0^2-4*0+6=6m x at 1 sec = 3m x at 2 sec = 2m x at 3 sec = 3m Now let's take out distance at particular second. x at 1st sec = 3m x at 2nd sec (x at 2 sec - x at 1sec) = 2-3 = 1m (as distance is scalar quantity and hence has no direction therefore no-ve sign) x at 3rd sec (x at 3 sec - x at 2sec) = 3-2 = 1m. Now add all the individual distances I.e. 3+1+1 = 5m.
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lllllll
• 0 • 1
F......k offf
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X denotes position (displacement) which can be negative but distance will always be positive hope my answer helps • -3
So here's the solution.
Hope you like it
A+B=74 • -4
May this is correct . I know not clear but try to under stand then understand • 1
9 metre
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The position of a • 0 • 1
Place t =0 u will get x =0
Similarly for t=3 u will get x=3
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Displacement 3m
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1??5 m is the ans
• 0 • 0
4.5 second
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Kfkgobpnjpmlljdai tk
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X =t - 4 + t b First we have to check where velocity of particle is change v= DX DT = 2+-4=2 (t-2) velocity is changing at t 2 sec motion of particle is like at t=0= X= 6so , magnitude of distance travelled (6-2)=4m at +=3=X=3 so distance travelled during t = 2sec of =X3 -x 2=3-2 =1m so Total distance travelled =(4+1)m = 5m
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Yoyo
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Thfhe
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3m option (4)
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Swag man
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tn nn
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Root(root3)+9
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I don't know plz hindu
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4 3meaters
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what means
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• 0
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Distance travelled by a body t=0s to t=3s is 3m
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2.7 into 10 power minus 7
• 0 • 0
The correct option is A.

Given,

x=t2?4t+6

So, the change in velocity is:

dtdx?=2t?4

Since velocity is changing,

At?t=0,x1?=6

At?t=2,x2?=2

So the magnitude of the distance traveled is:

6?2=4m

At?t=3,x3?=3

So distance traveled from?t=2s?to?t=3s

x3??x2?=3?2=1m

Thus the total distance is:

4+1=5m
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+50433710444
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